Root test: Question bank — Maths deep dive

Setup: everything this page assumes, from zero

We study an infinite sum $\sum a_{n} = a_{1} + a_{2} + a_{3} + \dots$ . "Converges" means the running totals settle on a single finite number; "diverges" means they don't.

Recall Quick reminders of the other tests named on this page

Comparison test: if $0 \leq ∣ a_{n} ∣ \leq b_{n}$ eventually and $\sum b_{n}$ converges, then $\sum ∣ a_{n} ∣$ converges. (Engine of the $L < 1$ case.) n-th term divergence test: if $a_{n} \neq \to 0$ , the series diverges. (Engine of the $L > 1$ case.) Ratio test: looks at $lim ∣ a_{n + 1} / a_{n} ∣$ instead of roots; gives the same $L$ when that limit exists. p-series rule: $\sum 1/ n^{p}$ converges iff $p > 1$ . (The standard $L = 1$ examples.)

The picture below shows the whole logic: extract $r$ by rooting, then judge against $1$ .

Every trap below attacks a specific soft spot: what $L = 1$ really means, why the proof needs $L < r < 1$ , how $n$ -th roots treat polynomials vs exponentials, and how the root test relates to the Ratio test.

True or false — justify

L = 1

then the series diverges.

False.

L = 1

means the test gives no information —

\sum 1/ n

diverges but

\sum 1/ n^{2}

converges, and both have

L = 1

. You must switch to another tool like the p-series rule.

L < 1

the series converges absolutely, not just conditionally.

True. The proof bounds

∣ a_{n} ∣ < r^{n}

and

\sum r^{n}

converges, so

\sum ∣ a_{n} ∣

converges — that is exactly absolute convergence.

The root test can prove conditional convergence (converges but not absolutely).

False. It only ever bounds

∣ a_{n} ∣

, so a "converges" verdict is always absolute convergence. For conditional cases (like the alternating harmonic series) you need the alternating series test.

L > 1

, the series diverges because its partial sums oscillate.

False in reasoning. It diverges because

∣ a_{n} ∣ > 1

infinitely often, so

a_{n} \neq \to 0

; the n-th term divergence test forces divergence regardless of oscillation.

n ∣ a_{n} ∣ \to 0

, the series still might diverge.

False.

L = 0 < 1

, and

0

is a perfectly valid case of

L < 1

, so it converges absolutely. Nothing special happens at

L = 0

The root test and Ratio test always agree on the same series.

False. They give the same

L

for many series, but the root test succeeds on some series where the ratio limit does not exist — the root test is strictly stronger.

If the ratio test is inconclusive (

L = 1

), the root test is also inconclusive.

True (they share the same $L$ ). Whenever the ratio limit exists it equals the root limit, so a

1

from one is a

1

from the other. The root test only wins when the ratio limit fails to exist at all.

A series with

L = \infty

still needs a separate divergence argument.

False.

L = \infty

is just an extreme case of

L > 1

; the same

n

-th term argument applies and the test itself concludes divergence.

Spot the error

"For

a_{n} = n^{2} / 2^{n}

n a_{n}

is a tiny number rooted, so

L \approx 0

Error. Polynomial factors root to

1

n n^{2} = n^{2/ n} \to 1

. Only the exponential survives, giving

L = 1/2

, not

0

n n \to 0

because you're taking a big root of a big number."

Error.

n n = e^{l n n / n}

and

ln n / n \to 0

, so the limit is

e^{0} = 1

, not

0

. The root grows much more slowly than

n

— it climbs toward

1

and stops there, never blowing up.

L > 1

means the terms eventually exceed

1

for all large

n

Error. With

lim sup

∣ a_{n} ∣ > 1

only needs to happen infinitely often, not for every large

n

. That is already enough to break

a_{n} \to 0

"Since any

r

with

L < r < 1

works, the choice of

r

makes the proof sloppy."

Error. Needing only one such

r

to dominate the tail is a strength; existence is guaranteed because

L < 1

leaves a gap. Freedom of choice does not weaken a logically valid step.

n ∣ a_{n} ∣ < r

holds for all

n

, so

∣ a_{n} ∣ < r^{n}

everywhere."

Error. The limit only guarantees the bound for

n \geq N

(eventually). The finitely many early terms don't matter for convergence, so we compare only the tail.

"To use the root test on

\sum \frac{3 ^{n}}{n ^{3}}

, I take

n 3^{n} / n^{3} = 3/ n^{3}

Error.

n n^{3} = (n n)^{3} \to 1

, not

n^{3}

. The correct root is

3/ (n n)^{3} \to 3

, giving

L = 3

(diverges).

L = 1

borderline, so I'll say it 'weakly converges'."

Error. There is no such verdict.

L = 1

is total silence — the test contributes nothing and you must use a different one.

Why questions

Why does the root test compare everything to a Geometric series and nothing else?

Because geometric series are the simplest family whose convergence we know exactly (

∣ r ∣ < 1

), and

n a_{n} \approx r

literally extracts the hidden geometric ratio from any term.

Why do we take the

n

-th root specifically, and not some other operation?

If a term behaves like

r^{n}

, the

n

-th root undoes the "

n

-times multiply" and leaves

r

bare. It is the exact inverse of raising to the

n

-th power, which is what geometric growth is.

Why is

lim sup

used in the definition instead of an ordinary limit?

The ordinary limit may not exist (the roots can wobble forever), but

lim sup

— the largest value they keep returning to — always exists (finite or

\infty

), so the test is always defined. For exam problems the plain limit usually exists and equals it.

Why does

L = 1

happen for every

p

\sum 1/ n^{p}

n 1/ n^{p} = e^{- p l n n / n}

and

ln n / n \to 0

, so the exponent vanishes for any fixed

p

. The root test literally cannot see the exponent

p

that decides convergence.

Why is the root test preferred over the Ratio test for terms like

(stuff)^{n}

The

n \cdot

cancels the outer power in one clean step, while the ratio test forces you to simplify a messy

(\dots)^{n + 1} / (\dots)^{n}

quotient.

Why does the

L > 1

case use the n-th term divergence test instead of comparison?

Comparison needs a convergent dominating series, which fails when terms grow. Instead we just note the terms don't go to

0

, which is a direct divergence criterion.

Why can't we conclude convergence merely from

a_{n} \to 0

a_{n} \to 0

is necessary but not sufficient —

\sum 1/ n

has terms going to

0

yet diverges. The root test demands the rate of shrinking be geometric, which is stronger.

Edge cases

What does the root test say about

\sum r^{n}

itself with

r = 1

n 1 = 1

for all

n

, so

L = 1

— inconclusive. And indeed

\sum 1

diverges, but the test can't tell; you see it directly from the

n

-th term test.

What happens for

\sum r^{n}

with

0 < r < 1

(the model case)?

n r^{n} = r

, so

L = r < 1

exactly, and the test confirms convergence. The root test is calibrated to be exact on the very family it compares to.

What if

a_{n} = 0

for some terms — does

n 0

break the test?

No.

n 0 = 0

, which only pulls values down; the

lim sup

handles zero terms fine and they never cause divergence.

What is

L

for

\sum \frac{1}{( l n n ) ^{n}}

and why is it a clean win?

n 1/ (ln n)^{n} = 1/ ln n \to 0

, so

L = 0 < 1

, converges. The perfect

n

-th power collapses instantly under the root — the ideal trigger for this test.

For a power series

\sum c_{n} x^{n}

, what does the root test become?

It becomes the Cauchy–Hadamard formula

R = 1/ lim sup n ∣ c_{n} ∣

— the root test applied with

x

folded in gives the Radius of convergence directly.

What if

n ∣ a_{n} ∣

oscillates between two values, say

0.4

and

0.9

lim sup

takes the largest recurring value, here

0.9 < 1

, so the series still converges. Using the ordinary limit would fail since it doesn't exist, which is exactly why

lim sup

is used.

Recall One-line summary of every trap

$L < 1$ ⇒ absolute convergence; $L > 1$ ⇒ divergence (terms not to $0$ ); $L = 1$ ⇒ silence. Polynomials root to $1$ , only exponentials survive, and $lim sup$ keeps the test always defined.